DT = (G 6 -Gd)-Y (44Z>)where DT = distance in Fig. 20.
Applying Eq. 42 with respect to station 53 + OO yields x = 40 ft (12.2 m); G =
0.0002(40) - 0.046 = -0.038; Gav = (-0.046 - 0.038)72 = -0.042; y = -0.042(40) = -1.68
ft (- 51.206 cm); elevation = 303.66 - 1.68 = 301.98 ft (9204.350 cm). Perform these cal-
culations for each station, and record the results in tabular form as shown:
Station jc, ft (m) G Gav y, ft (m) Elevation, ft (m)52 + 60 O (O) -0.046 -0.046 O (O) 303.66(92.56)
53 + 00 40(12.2) -0.038 -0.042 -1.68 (-0.51) 301.98(92.04)
53 + 50 90(27.4) -0.028 -0.037 -3.33 (-1.01) 300.33(91.54)
54 + 00 140(42.7) -0.018 -0.032 -4.48 (-1.37) 299.18(91.19)
54 + 50 190(57.9) -0.008 -0.027 -5.13 (-1.56) 298.53(90.99)
55 + 00 240(73.2) +0.002 -0.022 -5.28 (-1.61) 298.38(90.95)
55 + 50 290(88.4) +0.012 -0.017 -4.93 (-1.50) 298.73(91.05)
55 + 80 320(97.5) +0.018 -0.014 -4.48 (-1.37) 299.18(91.19)
- Verify the foregoing results
Apply the principle that for a uniform horizontal spacing the "second differences" be-
tween the ordinate are equal. The results are shown:
Calculation of Differences
Elevations, ft (m) First differences, ft (m) Second differences, ft (m)
301.98(92.04)
1.65(0.5029)
300.33(91.54) 0.50(0.1524)
1.15(0.3505)
299.18(91.19) 0.50(0.1525)
0.65(0.1981)
298.53(90.99) 0.50(0.1524)
0.15(0.0457)
298.38(90.95) 0.50(0.1524)
-0.35(0.10668)
298.73(91.05)- Apply the tangent-offset method to find the elevation
of each station
Since this method is based on Eq. 41, substitute directly in that equation. For the present
case, the equation becomes y = rx^2 /2 + G^x = 0.000 Ix^2 — 0.046*. Record the calculations
for y in tabular form. The results, as shown, agree with those obtained by the average-
grade method.